The ,locus of the point of intersection of two perpendicular tangents to the parabola `y^(2)=4ax` is

Options

A. x+a=0
B. x-a=0
C. y+a=0
D. y-a=0

Answer

To find the locus of the point of intersection of two perpendicular tangents to the parabola \( y^2 = 4ax \), we can follow these steps:

### Step 1: Equation of the Tangent
The equation of the tangent to the parabola \( y^2 = 4ax \) at a point \( (at^2, 2at) \) is given by:
\[ yy_1 = 2a(x + x_1) \]
where \( (x_1, y_1) = (at^2, 2at) \).

### Step 2: Find the Slopes of the Tangents
Let the slopes of the two tangents be \( m_1 \) and \( m_2 \). Since the tangents are perpendicular, we have:
\[ m_1 \cdot m_2 = -1 \]

### Step 3: Equation of the Tangents
The equations of the tangents can be expressed as:
\[ y = m_1x + c_1 \quad \text{and} \quad y = m_2x + c_2 \]
where \( c_1 \) and \( c_2 \) are the y-intercepts.

### Step 4: Intersection of the Tangents
To find the point of intersection of these tangents, we set the two equations equal to each other:
\[ m_1x + c_1 = m_2x + c_2 \]
Rearranging gives us:
\[ (m_1 - m_2)x = c_2 - c_1 \]
Thus, the x-coordinate of the intersection point is:
\[ x = \frac{c_2 - c_1}{m_1 - m_2} \]

### Step 5: Substitute for y
Substituting \( x \) back into one of the tangent equations to find \( y \):
\[ y = m_1\left(\frac{c_2 - c_1}{m_1 - m_2}\right) + c_1 \]

### Step 6: Eliminate Parameters
To find the locus, we need to eliminate the parameters \( m_1 \) and \( m_2 \). Using the condition \( m_1 \cdot m_2 = -1 \) and substituting \( m_2 = -\frac{1}{m_1} \) into the equations will help us derive a relationship between \( x \) and \( y \).

### Step 7: Derive the Locus Equation
After substituting and simplifying, we can derive the equation of the locus. The final result will be:
\[ x^2 + y^2 = 4a^2 \]
This represents a circle with center at the origin and radius \( 2a \).

### Conclusion
Thus, the locus of the point of intersection of two perpendicular tangents to the parabola \( y^2 = 4ax \) is:
\[ x^2 + y^2 = 4a^2 \]

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